I have the operator $Tf(x)=\int_0^x(x-y)f(y)dy$, for $f\in\mathcal{C}([0,1])$, equipped with the supremum/infinity norm. I know that such an operator is called a "Fredholm operator", and I am aware of several theorems that can be applied to it, for instance to show that this operator is compact.
I want to find the dimension of the range of this operator. Can I use compactness to say something about this? I've gone through my notes, and I can't find anything that could help me here.
Finding the dimension of the range of operator
compact-operatorsfunctional-analysis
Best Answer
We represent the operator in the form $Tf(x)=x\displaystyle\int\limits_0^xf(y)dy-\int\limits_0^xyf(y)dy$. It's easy to see that functions $x^n$, $n=2,3,4,...$ lies in the range of $T$, since the functions $n(n-1)x^{n-2}$ pass into them under the action of the $T$ (you can just solve the equation $Tf(x)=x^n$ by differentiating it twice to see this). Thus, in the range lies infinitely many linearly independent functions, therefore this image is infinite-dimensional.